• # question_answer A ray of light goes from medium 1 to medium 2, velocity of light in the two mediums are ${{C}_{1}}$ and ${{C}_{2}},$ respectively. For an angle of incidence, $\theta$ in medium 1. The corresponding angle of refraction in medium 2 is $\theta /2.$ (i) Which of the two media is optically denser and why? (ii) Establish the relationship among $\theta ,$ ${{c}_{1}}$ and ${{c}_{2}}.$ (iii) The critical angle of incidence in a glass slab placed in air $45{}^\circ .$ What will be the critical angle when it is immersed in water of refractive index 1.33? Or A beam of light of wavelength 400 nm is incident normally on a right angled prism as shown in the figure at night. It is observed that the light just grazes along the surface AC after falling on it. Given that the refractive index of the material of the prism varies with the wavelength $\lambda$ as per the relation,             ${{\mu }_{A}}=1.2+\frac{b}{{{\lambda }^{2}}}$ Right angled prism Calculate the value of b and the refractive index of the prism material for a wavelength $\lambda =5000\overset{{}^\circ }{\mathop{A}}\,.$ $[\text{given,}\theta =si{{n}^{-1}}(0.625)]$

(i) Smaller value of angle of refraction $\left( \frac{\theta }{2} \right)$ as compared to angle of incidence $(\theta )$ indicates bending of light towards the normal in second medium. Hence, medium 2 is optically denser. (ii) By Snell?s law, ${}_{1}{{\mu }_{2}}=\frac{\sin i}{\sin r}=\frac{\sin \theta }{\sin \theta /2}$             $=\frac{2\sin \theta /2\cos \theta /2}{\sin \theta /2}=2\cos \theta /2$ Also,     ${}_{1}{{\mu }_{2}}=\frac{{{c}_{1}}}{{{c}_{2}}}$ $\therefore$      $\frac{{{c}_{1}}}{{{c}_{2}}}=2\cos \theta /2$ or         $\theta =2{{\cos }^{-1}}\left( \frac{{{c}_{1}}}{2{{c}_{2}}} \right)$ (iii) ${}^{a}{{\mu }_{g}}=\frac{1}{\sin {{i}_{c}}}=\frac{1}{\sin 45{}^\circ }=\sqrt{2}=1.414$ Refractive index of glass w.r.t. water will be ${}^{w}{{\mu }_{g}}=\frac{{}^{a}{{\mu }_{g}}}{{}^{a}{{\mu }_{w}}}=\frac{1.414}{1.33}$ When glass slab is immersed in water, then critical angle $i{{'}_{c}}$ is given by $\sin i{{'}_{c}}=\frac{1}{{}^{w}{{\mu }_{g}}}=\frac{1}{\frac{1.414}{1.33}}=\frac{1.33}{1.414}=0.9432$$\therefore i{{'}_{c}}=70{}^\circ .36'$ Or Wavelength of incident light, $\lambda =400\,nm=400\times {{10}^{-9}}m=4\times {{10}^{-7}}m$ For refracting surface AB, ${{i}_{1}}=0{}^\circ ,$ ${{r}_{1}}=0{}^\circ$ For refracting surface AC, ${{r}_{2}}=90{}^\circ$ and ${{i}_{2}}=\theta$ Using Snell's law at surface AC, $\frac{\sin {{i}_{2}}}{{{\operatorname{sinr}}_{2}}}=\frac{1}{\mu }\Rightarrow \frac{\sin \theta }{\sin 90{}^\circ }=\frac{1}{\mu }$ $\Rightarrow$   $\frac{1}{\sin \theta }=\mu$     $[\because \sin 90{}^\circ =1]$ $\Rightarrow$   $\mu =\frac{1}{0.625}=1.6$     $[\text{given},\sin \theta =0.625]$ But the relation is $\mu =12+\frac{b}{{{\lambda }^{2}}}$        [given] $\therefore$      $1.6=12+\frac{b}{{{(4\times {{10}^{-7}})}^{2}}}$ $b=0.4\times {{(4\times {{10}^{-7}})}^{2}}=6.4\times {{10}^{-14}}{{m}^{2}}$ For wavelength, $\lambda =5000\,\overset{{}^\circ }{\mathop{A}}\,=5\times {{10}^{-7}}m$ Refractive index, $\mu =1.2+\frac{6.4\times {{10}^{-14}}}{{{(5\times {{10}^{-7}})}^{2}}}=1.2+\frac{6.4}{25}$ $=1.2+0.256=1.456$
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